Describing the path of a heat seeking particle

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SUMMARY

The discussion focuses on determining the path of a heat-seeking particle starting at the coordinates (4,3,10) within a temperature field defined by the equation T(x,y,z) = 400 - 2x² - y² - 4z². The gradient of the temperature function, ∇T = < -4x, -2y, -8z >, indicates the direction of maximum temperature increase. The particle's velocity vector is proportional to the gradient, leading to the system of differential equations dx/dt = -4kx, dy/dt = -2ky, and dz/dt = -8kz. By solving these equations with the initial conditions, parametric equations for the particle's path can be derived.

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Homework Statement



Find the path of a heat-seeking particle placed at (4,3,10) with a temperature field

[tex]T(x,y,z) = 400 - 2x^2 - y^2 -4z^2[/tex]

Homework Equations



Formulas for directional derivative and gradient.

The Attempt at a Solution



At any point in space (x,y,z), the particle must be moving in the direction which causes the greatest increase in T. This direction is given by the direction of

[tex]\nabla T = < -4x, -2y, -8z >[/tex].

I do not know how to continue from here.
 
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The velocity vector points in the direction of [itex]\nabla T[/itex] and so must be some multiple, say "k", of it.

That means you must have dx/dt= -4kx, dy/dt= -2ky, and dz/dt= -8kz for some number k.
You also are given that x(0)= 4(0), y(0)= 3, z(0)= 10. You can solve each of those for x, y, and z in terms of kt. That gives parametric equations for the path with parameter s= kt.
 

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